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Full text of "Treatise On Analysis Vol-Ii"

420       XV   NORMED ALGEBRAS AND SPECTRAL THEORY

ing results then imply that (N - x \ N  y) = (N* -x\N*-y) for all x, y in
dom(N)= dom(JV*) = F. This being so, if z e domC/WV*), then we have

(N  x\ N- z) = (N* - x \N* - z) = (x | NN*  z)

for all x e dom(A^). This proves that N  z e dom(JV*), by the definition of the
adjoint (15.12.3), and that N*N - z = NN* - z. Thus we have established that
dom(NN*) <=. dom(N*N) and that the operators NN* and N*N coincide
on dom(7\W*). The proof of (ii) is now completed by interchanging the roles
of AT and N* (since N** = N).

Now return to the proof of (i). Consider the continuous hermitian
operator B = (/ -f N*N)~"1 (15.12.6), whose spectrum is contained in I = [0, 1].
We shall first show that

f(B)(dom(NJ) c domCAO       and      f(B)N - x = Nf(B)  x

for all/6 ^C(I) and all x e dom(N). First of all take/= lc; if x e dom(AO,
we may write

BN'X-= BN(I + N*N)B * x = JS(N + NN*N)B  x,

recalling that NB(E) c dom(A^*) and N*NB -x = x-B-xe dom(N), be-
cause fl(E) c dom(^) (15.12.6). But N*N = NN*, hence

BN*x=* B(N + N*NN)B * x = B(1 4- N*N)NB -x = NB-x;

that is, fl/V"  ^r = 7V#  A: for all x e dom(N). By induction on n, we obtain
J"(dom(JV)) a dom(N) &nd BnN - x = NBn  x for all jc e dom(N) and all
integers n > 0. This establishes our assertion when /is a polynomial with real
coefficients. Hence, for such a polynomial / we have

(15.12.8.2)              (f(B)N -x \y) = (/()  x | N*  y)

for all x e dom(AO and j e dom(AO.

Having regard to (15.11.2) applied to B, we deduce that

J/(0 dm*.,.,(0 = J

/(0 dm,,^.,(0,

and since the restrictions of polynomials with real coefficients to Sp(2?)
are dense in ^R(Sp(J)) by the Weierstrass approximation theorem (7.4.1 ), the
formula above shows that mN.Xty = mX)N*.y ((3.15.1) and (13.2)). Conse-
quently, by applying (15.11.2) to B, the formula (15.112,8.) is valid for all
functions /e^c(I). But this formula shows that f(B)  x e dom(7V**) =
dom(AT) and that (f(B)N -x\y) = (Nf(B) -x\y) for all y e dom(N). Since
dom(AT) is dense in E, it follows that Nf(B) 'X=f(B)N-x for all x e dom(N).                        n